Klein-Gordon equation

How to construct

Starting from the relativistic connection between energy and momentum:

$E^{2}={\mathbf {p}}^{2}c^{2}+m^{2}c^{4}$

Substituting $E\rightarrow i\hbar {\frac {\partial }{\partial t}}$ and ${\mathbf {p}}\rightarrow -i\hbar \nabla$, we get Klein-Gordon equation for free particles as follows:

$-\hbar ^{2}{\frac {\partial ^{2}\psi ({\mathbf {r}},t)}{\partial t^{2}}}=(-\hbar ^{2}c^{2}\nabla ^{2}+m^{2}c^{4})\psi ({\mathbf {r}},t)\qquad \qquad \qquad \qquad \qquad (1)$

Klein-Gordon can also be written as the following:

$(\square -K^{2})\psi ({\mathbf {r}},t)=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;\;(2)$

where $\square =\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}$ is d'Alembert operator and $K={\frac {mc}{\hbar }}$.

Equation (2) looks like a classical wave equation with an extra term $K^{2}$.

For a charged particle couple with electromagnetic field, Klein-Gordon equation is as follows:

$\left[i\hbar {\frac {\partial }{\partial t}}-e\phi ({\mathbf {r}},t)\right]^{2}\psi ({\mathbf {r}},t)=\left(\left[-i\hbar \nabla -{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)\right]^{2}c^{2}+m^{2}c^{4}\right)\psi ({\mathbf {r}},t)$

Klein-Gordon is second order in time. Therefore, to see how the states of a system evolve in time we need to know both $\psi ({\mathbf {r}},t)$ and ${\frac {\partial \psi ({\mathbf {r}},t)}{\partial t}}$ at a certain time. While in nonrelativistic quantum mechanics, we only need $\psi ({\mathbf {r}},t)$

Also because the Klein-Gordon equation is second order in time, it has the solutions $\psi ({\mathbf {r}},t)=e^{i({\mathbf {p}}{\mathbf {r}}-Et)/\hbar }$ with either sign of energy $E=\pm c{\sqrt {{\mathbf {p}}^{2}+m^{2}c^{2}}}$. The negative energy solution of Klein-Gordon equation has a strange property that the energy decreases as the magnitude of the momentum increases. We will see that the negative energy solutions of Klein-Gordon equation describe antiparticles, while the positive energy solutions describe particles.

Continuity equation

Multiplying (1) by ${\mathbf {\psi }}^{*}$ from the left, we get:

$-{\frac {\hbar ^{2}}{c^{2}}}\psi ^{*}{\frac {\partial ^{2}\psi ({\mathbf {r}},t)}{\partial t^{2}}}=\psi ^{*}(-\hbar ^{2}\nabla ^{2}+m^{2}c^{2})\psi ({\mathbf {r}},t)\qquad \qquad \qquad (3)$

Multiplying the complex conjugate form of (1) by ${\mathbf {\psi }}$ from the left, we get:

$-{\frac {\hbar ^{2}}{c^{2}}}\psi {\frac {\partial ^{2}\psi ^{*}({\mathbf {r}},t)}{\partial t^{2}}}=\psi (-\hbar ^{2}\nabla ^{2}+m^{2}c^{2})\psi ^{*}({\mathbf {r}},t)\qquad \qquad \qquad (4)$

Subtracting (4) from (3), we get:

$-{\frac {\hbar ^{2}}{c^{2}}}\left(\psi ^{*}{\frac {\partial ^{2}\psi }{\partial t^{2}}}-\psi {\frac {\partial ^{2}\psi ^{*}}{\partial t^{2}}}\right)=\hbar ^{2}\left(\psi \nabla ^{2}\psi ^{*}-\psi ^{*}\nabla ^{2}\psi \right)$

$\Rightarrow -{\frac {\hbar ^{2}}{c^{2}}}{\frac {\partial }{\partial t}}\left(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}}\right)+\hbar ^{2}\nabla \left(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*}\right)=0$

$\Rightarrow {\frac {\partial }{\partial t}}\left[{\frac {i\hbar }{2mc^{2}}}\left(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}}\right)\right]+\nabla \left[{\frac {\hbar }{2mi}}\left(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*}\right)\right]=0$

this give us the continuity equation:

${\frac {\partial \rho }{\partial t}}+\nabla {\mathbf {j}}=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (5)$

where $\rho ={\frac {i\hbar }{2mc^{2}}}\left(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}}\right)\qquad \qquad \qquad \qquad \qquad \;\;\;\;(6)$

${\mathbf {j}}={\frac {\hbar }{2mi}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})\qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;(7)$

From (5) we can see that the integral of the density ${\mathbf {\rho }}$ over all space is conserved. However, ${\mathbf {\rho }}$ is not positively definite. Therefore, we can neither interpret ${\mathbf {\rho }}$ as the particle probability density nor can we interpret ${\mathbf {j}}$ as the particle current. The appropriate interpretation are charge density for $e\rho ({\mathbf {r}},t)$ and electric current for $e{\mathbf {j}}({\mathbf {r}},t)$ since charge density and electric current can be either positive or negative.

Nonrelativistic limit

In nonrelativistic limit when $v\ll c$ or $p\ll mc$, we have:

$E=[(pc)^{2}+(mc^{2})^{2}]^{1/2}=mc^{2}\left[1+({\frac {p}{mc}})^{2}\right]^{1/2}\approx mc^{2}\left(1+{\frac {1}{2}}\left({\frac {p}{mc}}\right)^{2}\right)=mc^{2}+{\frac {p^{2}}{2m}}$

So, the relativistic energy is different from classical energy by $mc^{2}$, therefore, we can expect that if we write the solution of Klein-Gordon equation as $\psi e^{-imc^{2}t/\hbar }$ and substitute it into Klein-Gordon equation, we will get Schrodinger equation for ${\mathbf {\psi }}$.

Indeed, doing so we get:

$-\hbar ^{2}{\frac {\partial ^{2}\psi }{\partial t^{2}}}e^{-imc^{2}t/\hbar }+2{\frac {\partial \psi }{\partial t}}imc^{2}\hbar e^{-imc^{2}t/\hbar }+\psi m^{2}c^{4}e^{-imc^{2}t/\hbar }=-\hbar ^{2}c^{2}\nabla ^{2}\psi e^{-imc^{2}t/\hbar }+m^{2}c^{4}\psi e^{-imc^{2}t/\hbar }$

$\Rightarrow -{\frac {\hbar ^{2}}{2mc^{2}}}{\frac {\partial ^{2}\psi }{\partial t^{2}}}+i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi$

In the nonrelativistic limit the first term is considered negligibly small. As a result, for free particles in this limit we get back the Schrodinger equation:

$i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi$